Potassium superoxide, KO2, is often used in oxygen masks (such as those used by firefighters) because KO2 reacts with CO2 to release molecular oxygen. Experiments indicate that 2 mol of KO21s2 react with each mole of CO21g2. (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?

Video transcript

Hey everyone, we're asked to give the oxidation number for each element. Starting off with our react inside. Let's look at our platinum for sulfate. So we know that our platinum has a charge of plus four and that's because we require to sulfates in order to balance out that plus four. So platinum has an oxidation state of plus four And we know that oxygen has an oxidation state of -2. And if we solve for X and say sulfate is X, we can simply do X plus four times negative two since we have four oxygen's And we can equal this to negative two since the overall charge of sulfate is -2, solving for X, we get x minus eight equals negative two and X will be plus six. So silver's oxidation state is going to be plus six. Moving on to rubidium iodine, we know that rubidium has an oxidation state of plus one. Since it is a group won a medal, iodine has an oxidation state of -1 since it's a group seven a medal. Looking at our product side, we can see that platinum has now become Plus two in its oxidation state since we have two. I'd eines balancing out our platinum and speaking of iodine, we can see that iodine ended up with an oxidation state of zero since it is now in its standard state. Looking at our sulfur and oxygen's, They still remained plus six and -2 respectively, and rubidium also remained A Plus one Oxidation State. So I hope that made sense. And let us know if you have any questions